Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;

int M;
int p[200001];
int now[200001];
int T[200001];
int f[200001];
int main(){
    cin>>M;
    REP(i,M){
       cin>>p[i];
       now[p[i]]=i;
    }
    T[M]=0;
    for(int i=M-1;i>=1;i--){

        if(T[i+1]==0){
           //cout<<i+1<<"aa "<<i<<endl;
           if(now[i+1]<now[i]){
              T[i]=1;
              f[i]=i+1;
           }else{
              T[i]=0;
              f[i]=i;
           }
        }else{
           //cout<<now[f[i+1]]<<" "<<now[i]<<" "<<now[i+1]<<endl;
           if((now[f[i+1]]<now[i]&&now[i]<now[i+1])||
              (now[i]<now[i+1]&&now[i+1]<now[f[i+1]])||
              (now[i+1]<now[f[i+1]]&&now[f[i+1]]<now[i])){
              T[i]=T[i+1];
              f[i]=f[i+1];
           }else{
              T[i]=T[i+1]+1;
              if(now[i]<now[i+1]){
                 f[i]=i+1;
              }else{
                 f[i]=i+1;
              }
           }
        }
        //cout<<now[f[i]]<<now[f[i+1]]<<endl;
    }
    
    
    cout<<T[1]<<endl;
    
    return(0);
}